As shown in the figure, the mass of the shell of the spring scale is M0, and the mass of the spring and the hook is ignored. The mass of a heavy object suspended by the hook is m. now the external force F in a vertical direction is used to pull the spring scale to make it move upward uniformly and accelerate. The reading of the spring scale is: () A. Mg; B. [M / (M0 + m)] mg; C. [M0 / (M0 + m)] f; D. [M / (M0 + m)] f according to Newton's second law Overall study f = (M0 + m) a Research object f bullet = ma The result shows that the reading F of spring scale is [M / (M0 + m)] f, why f = (M0 + m) a instead of F = (M0 + m) a + (M0 + m) g

As shown in the figure, the mass of the shell of the spring scale is M0, and the mass of the spring and the hook is ignored. The mass of a heavy object suspended by the hook is m. now the external force F in a vertical direction is used to pull the spring scale to make it move upward uniformly and accelerate. The reading of the spring scale is: () A. Mg; B. [M / (M0 + m)] mg; C. [M0 / (M0 + m)] f; D. [M / (M0 + m)] f according to Newton's second law Overall study f = (M0 + m) a Research object f bullet = ma The result shows that the reading F of spring scale is [M / (M0 + m)] f, why f = (M0 + m) a instead of F = (M0 + m) a + (M0 + m) g

Your understanding is correct
But the result is the same whether gravity is taken into account or not
F = (M 0 + m) a + (M 0 + m) g = (M 0 + m) (a + G), which is the same form as F = (M 0 + m) a
The two connectors gain power in proportion to their mass